Laurence Kirby and Jeff Paris introduced a graph theoretic hydra game with behavior similar to that of Goodstein sequences: the "Hydra" is a rooted tree, and a move consists of cutting off one of its "heads" (a branch of the tree), to which the hydra responds by growing a finite number of new heads according to certain rules. Kirby and Paris proved that the Hydra will eventually be killed, regardless of the strategy that Hercules uses to chop off its heads, though this may take a very, very, very long time.[1]

Goodstein sequences are defined in terms of a concept called "hereditary base-n notation". This notation is very similar to usual base-n positional notation, but the usual notation does not suffice for the purposes of Goodstein's theorem.

In ordinary base-n notation, where n is a natural number greater than 1, an arbitrary natural number m is written as a sum of multiples of powers of n:

where each coefficient satisfies , and . For example, in base 2,

Thus the base 2 representation of 35 is 100011, which means Similarly, 100 represented in base 3 is 10201:

Note that the exponents themselves are not written in base-n notation. For example, the expressions above include and .

To convert a base-n representation to hereditary base n notation, first rewrite all of the exponents in base-n notation. Then rewrite any exponents inside the exponents, and continue in this way until every digit appearing in the expression is n or less.

For example, while 35 in ordinary base-2 notation is , it is written in hereditary base-2 notation as

The Goodstein sequenceG(m) of a number m is a sequence of natural numbers. The first element in the sequence G(m) is m itself. To get the second, G(m)(2), write m in hereditary base 2 notation, change all the 2s to 3s, and then subtract 1 from the result. In general, the term G(m)(n+1) of the Goodstein sequence of m is as follows: take the hereditary base n+1 representation of G(m)(n), and replace each occurrence of the base n+1 with n+2 and then subtract one. Note that the next term depends both on the previous term and on the index n. Continue until the result is zero, at which point the sequence terminates.

Early Goodstein sequences terminate quickly. For example, G(3) terminates at the sixth step:

Base

Hereditary notation

Value

Notes

2

3

Write 3 in base 2 notation

3

3

Switch the 2 to a 3, then subtract 1

4

3

Switch the 3 to a 4, then subtract 1. Now there are no more 4s left

5

2

No 4s left to switch to 5s. Just subtract 1

6

1

No 5s left to switch to 6s. Just subtract 1

7

0

No 6s left to switch to 7s. Just subtract 1

Later Goodstein sequences increase for a very large number of steps. For example, G(4) A056193 starts as follows:

Hereditary notation

Value

4

26

41

60

83

109

253

299

Elements of G(4) continue to increase for a while, but at base , they reach the maximum of , stay there for the next steps, and then begin their first and final descent.

The value 0 is reached at base . (Curiously, this is a Woodall number: . This is also the case with all other final bases for starting values greater than 4.[citation needed])

However, even G(4) doesn't give a good idea of just how quickly the elements of a Goodstein sequence can increase. G(19) increases much more rapidly, and starts as follows:

Hereditary notation

Value

19

7,625,597,484,990

In spite of this rapid growth, Goodstein's theorem states that every Goodstein sequence eventually terminates at 0, no matter what the starting value is.

Goodstein's theorem can be proved (using techniques outside Peano arithmetic, see below) as follows: Given a Goodstein sequence G(m), we construct a parallel sequence P(m) of ordinal numbers which is strictly decreasing and terminates. Then G(m) must terminate too, and it can terminate only when it goes to 0. A common misunderstanding of this proof is to believe that G(m) goes to 0 because it is dominated by P(m). In fact that P(m) dominates G(m) plays no role at all. The important points is: G(m)(k) exists if and only if P(m)(k) exists (parallelism). Then if P(m) terminates, so does G(m). And G(m) can terminate only when it comes to 0.

More precisely, each term P(m)(n) of the sequence P(m) is obtained by applying a function f on the term G(m)(n) of the Goodstein sequence of m as follows: take the hereditary base n+1 representation of G(m)(n), and replace each occurrence of the base n+1 with the first infinite ordinal number ω. For example and . Addition, multiplication and exponentiation of ordinal numbers are well defined.

The base-changing operation of the Goodstein sequence when going from G(m)(n) to G(m)(n+1) does not change the value of f (that's the main point of the construction), thus after the minus 1 operation, P(m)(n+1) will be strictly smaller than P(m)(n). For example, , hence is strictly greater than

If the sequence G(m) did not go to 0, it would not terminate and would be infinite (since G(m)(k+1) would always exist). Consequently, P(m) also would be infinite (since in its turn P(m)(k+1) would always exist too). But P(m) is strictly decreasing and the standard order < on ordinals is well-founded, therefore an infinite strictly decreasing sequence cannot exist, or equivalently, every strictly decreasing sequence of ordinals does terminate (and cannot be infinite). This contradiction shows that G(m) terminates, and since it terminates, goes to 0 (by the way, since there exists a natural number k such that G(m)(k)=0, by construction of P(m) we have that P(m)(k)=0).

While this proof of Goodstein's theorem is fairly easy, the Kirby–Paris theorem which says that Goodstein's theorem is not a theorem of Peano arithmetic, is technical and considerably more difficult. It makes use of countable nonstandard models of Peano arithmetic. What Kirby showed is that Goodstein's theorem leads to Gentzen's theorem, i.e. it can substitute for induction up to ε0.

Suppose the definition Goodstein sequence is changed so that instead of replacing each occurrence of the base b with b+1 it was replaces it with b+2. Would the sequence still terminate? More generally, let be any sequences of integers. Then let the term G(m)(n+1) of the extended Goodstein sequence of m be as follows: take the hereditary base representation of G(m)(n), and replace each occurrence of the base with and then subtract one. The claim is that this sequence still terminates. The extended proof defines as follows: take the hereditary base representation of , and replace each occurrence of the base with the first infinite ordinal number. The base-changing operation of the Goodstein sequence when going from G(m)(n) to G(m)(n+1) still does not change the value of f. For example, if and if , then , hence the ordinal is strictly greater than the ordinal

The Goodstein function, , is defined such that is the length of the Goodstein sequence that starts with n. (This is a total function since every Goodstein sequence terminates.) The extreme growth-rate of can be calibrated by relating it to various standard ordinal-indexed hierarchies of functions, such as the functions in the Hardy hierarchy, and the functions in the fast-growing hierarchy of Löb and Wainer:

Kirby and Paris (1982) proved that

has approximately the same growth-rate as (which is the same as that of ); more precisely, dominates for every , and dominates

(For any two functions , is said to dominate if for all sufficiently large .)

Cichon (1983) showed that

where is the result of putting n in hereditary base-2 notation and then replacing all 2s with ω (as was done in the proof of Goodstein's theorem).

Goodstein's theorem can be used to construct a total computable function that Peano arithmetic cannot prove to be total. The Goodstein sequence of a number can be effectively enumerated by a Turing machine; thus the function which maps n to the number of steps required for the Goodstein sequence of n to terminate is computable by a particular Turing machine. This machine merely enumerates the Goodstein sequence of n and, when the sequence reaches 0, returns the length of the sequence. Because every Goodstein sequence eventually terminates, this function is total. But because Peano arithmetic does not prove that every Goodstein sequence terminates, Peano arithmetic does not prove that this Turing machine computes a total function.